$(6)\oplus(7)\ $ gives $\ \tag{6'}1=y\oplus z$
so that from $(8)$ $w=0$
from $(7)$ and $(6)$ $\ x=z$ and $\ \overline{x}=y$
from $(5)$ $\ \overline{x}=1$ and the final conclusion :
$$(w,x,y,z)=(0,0,1,0)$$
as found (earlier) by martini...

In this field, addition corresponds to the XOR operation, while multiplication corresponds to AND (as it does in the reals, if the operands are restricted to $0$ and $1$). As both $0$ and $1$ are their own additive inverses in $GF(2)$ (since $0 \oplus 0 = 1 \oplus 1 = 0$), subtraction is also equivalent to XOR, while division is trivial (dividing by $1$ does nothing, dividing by $0$ is undefined).

and $1\oplus 1=0$, so you get $z+w=0$. Substitue this into the last two equations to get $0=x\oplus 0=x$ and $1=y\oplus 0=y$. Now you know that $x=0$ and $y=1$ so $x\oplus y=1$. Substituting this into the first two equations, we find that $1=1\oplus z$ and $1=1\oplus w$. Add $1$ to both sides to get $0=z$ and $0=w$. The solution is therefore $x=0,y=1,z=0,w=0$.